Find the equation of a straight line whose y intercept is $- 4$ and is parallel to the line joining the points $(3, - 2)$ and $(1, 4)$ .

$x + 3 y + 4 = 0$

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$3 x + y - 3 = 0$

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$x + 3 y - 3 = 0$

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$3 x + y + 4 = 0$

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Solution

The correct option is D
$3 x + y + 4 = 0$

An equation of the straight line whose slope is $m$ and intercept on the y-axis is $c$

$y = m x + c$

Slope of required line is same as the slope of the parallel line

Slope parallel line $m$ = $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

= $\frac{4 + 2}{1 - 3}$

= $\frac{6}{- 2}$

= $- 3$

Intercept $c = - 4$

Equation of the straight line

$y = m x + c$

$y = - 3 x - 4$

$3 x + y + 4 = 0$

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Find the equation of a straight line whose y intercept is −4 and is parallel to the line joining the points (3,−2) and (1,4).

Find the equation of a straight line whose y intercept is $- 4$ and is parallel to the line joining the points $(3, - 2)$ and $(1, 4)$ .